Question

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8547 g and a standard deviation of 0.0511 g. A sample of these candies came from a package containing 463 ​candies, and the package label stated that the net weight is 395.2 g.​ (If every package has 463 ​candies, the mean weight of the candies must exceed StartFraction 395.2 Over 463 EndFraction 395.2 463equals=0.8535 g for the net contents to weigh at least 395.2 ​g.) a. If 1 candy is randomly​ selected, find the probability that it weighs more than 0.8535 g. The probability is .​(Round to four decimal places as​ needed.)

Answer 1

There is a 69.15% probability that it weighs more than 0.8535 g.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by

`Z = (X - \mu)/(\sigma)`

After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.

In this problem, we have that

The weights of a certain brand of candies are normally distributed with a mean weight of 0.8547 g, so
`\mu = 0.8547`.

We have a sample of 463 candies, so we have to find the standard deviation of this sample to use in the place of
`\sigma` in the Z score formula. We can do this by the following formula:

`s = (\sigma)/(√(463)) = 0.0024`

Find the probability that it weighs more than 0.8535

This is 1 subtracted by the pvalue of Z when
`X = 0.8535`

So

`Z = (X - \mu)/(s)`

`Z = (0.8535 - 0.8547)/(0.0024)`

`Z = -0.5`

`Z = -0.5` has a pvalue of 0.3085.

This means that there is a 1-0.3085 = 0.6915 = 69.15% probability that it weighs more than 0.8535 g.